Lecture 14 Matrix representing of Operators While representing operators in terms of matrices, we use the basis kets to compute the matrix elements of the operator as shown below < Φ 1 x Φ 1 >< Φ 1 x Φ > < Φ 1 x Φ n > X = ( < Φ n x Φ 1 >. < Φ n x Φ n >) Where Φ n > s are the eigenkets of X The expectation value of X is < X > =< Φ X Ψ > = n,m < Ψ Φ n >< Φ n x Φ n >< Φ m Ψ > If two operators A, B commute then they have same set of eigenkets. Change of basis Suppose we have t observables A and B. The ket space is spanned by { Φ >} and { x >}. e.g. for a spin 1 system, we can have a description in terms of s z±> or s x ±> or s y ±> where s i, (i x, y, z) are the components of the spin 1 matrices. Either of these different set of base kets span the same space. We are interested in finding out how these two descriptions are related. Changing the set of base kets is referred to as a change of basis or a change of representation. Thus given two sets of basis kets, both satisfying orthonormality and completeness, there exists an unitary operator U such that { x >} = U{ Φ >} Where U is an unitary operator satisfying, UU = 1 and U U = 1 Eigenvectors corresponding to different eigenvalues are linearly independent. Let u, v are eigenvectors of the matrix A corresponding to λ 1 and λ, then au + bv = 0 if they are not independent. Opearating by A. a, b 0. aλ 1 u + bλ u = 0 Joint initiative of IITs and IISc Funded by MHRD Page of 8
(1) and () have unique solutions if Det a b = 0 ab(λ aλ 1 bλ λ 1 ). Thus a, b = 0 Hence they are linearly independent. Summarizing, a change of basis kets is an unitary operation that establishes the link between the two kets. Linear Algebra in Quantum Mechanics We have seen in quantum mechanics that the state of an electron subjected to some potential is given by a wave function Ψ(r, t) which is obtained from the solution of a linear second order differential equation, called the Schrodinger equation. This implies that if Ψ 1 and Ψ are the solutions for the Schrodinger equation, then C 1 Ψ 1 + C Ψ is also a solution of the equation with C 1 and C as arbitrary complex numbers. It is actually the linearity of Schrodinger equation that allows for the superposed solution as a valid solution. The commutativity, associativity for Ψ are trivially satisfied. Another close link between the content of this chapter to quantum mechanics can be mentioned as follows. Since all the physical observables correspond to real values, they must correspond to Hermitian matrices. As an example, the students can be asked that operator like the momentum (P x = ih ) or the Pauli matrix σ x y ( 0 i i 0 ) which contain i(= 1), however are still unitary and yield real eigenvalues. Also for another example, one can show that eigenvectors of σ y corresponding to the (real) eigenvalues (±1) are orthogonal. Consider a basis set {u i } for (i = 1,. n) in a R n space. The set {u i } is orthonormal if < u i u j > = α 3 r u i (r )u j (r ) = δ ij Thus every function Ψ(r ) R n can be expanded in one (and only one) way in terms of u i (r ) Ψ(r ) = i c i u i (r ) which means that the set c i is unique and can be found by, c i = < u i Ψ > = d 3 r u i (r )Ψ(r ). Let Φ(r ) and Ψ(r ) be two functions which are expand using the basis {u i } Φ(r ) = i b i u i (r ) Ψ(r ) = i c j u j (r ) The inner product of these two functions is expressed as, < Φ Ψ > = b i i c i and < Ψ Ψ > = i c i. Joint initiative of IITs and IISc Funded by MHRD Page 3 of 8
Change of basis For a plane wave, Ψ(x) = 1 Ψ (p) = 1 πh πh dpψ (p)e ipx h dxψ(x)e ipx h < Ψ(x) Ψ(x) > = < Ψ (p) Ψ (p) > This is known as Perseval s identity. Projection Operators A projection operator p is a self adjoint operator satisfying the condition, p = p Let p 1 and p be two projection operators, then the question is whether p 1 + p is also a projection operator. (p 1 + p ) = p 1 + p + p 1 p + p p 1 p 1 + p is also a projection operator if p 1 p + p p 1 = 0 (1) That is if they anticommute. The relation between p 1 and p is more clearly expressed by left multiplying (1) by p 1, p + p 1 p p 1 = 0 () Again right multiplying by p 1 p 1 p p 1 + p = 0 (3) Subtracting () and (3) p 1 p p p 1 = 0 (4) Thus (4) and (1) are simultaneously satisfied if p 1 p = 0 = p p 1. Thus the necessary and sufficient condition are that the two operators have to be mutually orthogonal. Some important properties of Hermitian Matrices 1. Two Hermilian matrices can be simultaneously diagonalised if and only if they commute. The converse is also true.. The angle of rotation θ for the rotation described by a proper orthogonal matrix R θ is given by, cosθ = tr[r θ ] 1 3. For a matrix: a) All eigenvalues λ are distinct. There are three mutually orthogonal eigenvectors. b) Two of the eigenvalues are equal λ 1 = λ. The corresponding eigenvector is unique. But one can always generate an arbitrary orthogonal eigenvector. Joint initiative of IITs and IISc Funded by MHRD Page 4 of 8
Thus these two along with the eigenvector corresponding to λ 3 constitutes the three eigenvectors. c) λ 1 = λ = λ 3 any three mutually perpendicular eigenvectors will do. To summarize this chapter, a little more can be said about the Hermitian matrices and their relevance to quantum mechanics. To prove that the eigenvalues as real, one may proceed in the following manner. A α >= a α > Inner product with α > yields < α A α >= a < α α > The adjoint equation can be denoted as, < α A = a < α (remembering A = A ) Again taking an inner product with α > yields the result a = a, or in other words, the eigenvalues of Hermilian matrices are real. Likewise, the eigenkets corresponding to different eigenvalues are orthogonal. A α >= a α > A β >= β β > < α A β >= β < α β > < β A α >= α < β α > Because of the Hermilian nature of the eigenvalues, < α A β >= β A α > Hence (α β) < α β > = 0 Thus if α β < α β > = 0 So they are orthogonal. Further eigenvectors of a Hermitian matrix span the space. Joint initiative of IITs and IISc Funded by MHRD Page 5 of 8
Tutorials 1. Prove each of the following, a) P is orthogonal if and only if P T is orthogonal b) If P is orthogonal, then P 1 is orthogonal. c) If P and Q are orthogonal, then PQ is orthogonal. Ans: a) (P T ) T = P. Thus P is orthogonal if and only if PP T = I if and only if (P T ) T P T = I, that is, if and only if P T is orthogonal. b) We have P T = P 1 since P is orthogonal. Thus by part (a) P 1 is orthogonal. c) We have P T = P 1, Q T = Q 1. Prove that momentum operator in quantum mechanics is a Hermitian Operator. < p x > = Ψ ( ih ) Ψdx x = - ih Ψ Ψ dx x = -iħ [Ψ Ψ Ψ(x) dψ dx] 0 = iħ Ψ(x) dψ dx = < p x > dx dx 3. A particle of mass m is in a one-dimensional box of width a. At t = 0, the particle is in the state Ψ(x, 0) = 3Φ +4Φ 9, The Φ 5 n function are orthogonal eigenstates of the Hamitian HΦ n = a sin ( πx a) what will be the measurement of energy, E yield at z = 0. What is the probability of finding this value? Ans: First check that Ψ is properly normalized. < Ψ Ψ > = 1 Ψ = n c n Φ n >, c = 3, c 5 9 = 4, c 5 n = 0 for n or 9. P(E ) = 9 and P(E 5 9) = 16. 5 Joint initiative of IITs and IISc Funded by MHRD Page 6 of 8
In an ensemble of 500 identical one-dimensional boxes each containing an identical particle in the same state Ψ(x, 0). Measurement E at t = 0, finds 900 particle of each E = 4E, and 1600 particles E 9 = 81E. 4. It can be seen that a, 1 a and a 3 are all linearly independent as they do not lie in a plane. Thus, a unit vector in the direction of a 1 is v 1 = a 1 = 1 a 1 3 [, 3, 0] = [0.554, 0.831, 0]. We subtract from a, the component of a in the direction of v. 1 v = a < v a 1 > u 1 < v a 1 > = 4.16 < v a 1 > v 1 = [.30, 3.45, 0] Hence, v = [3.7,.45, 0]. Hence calculate the unit vector in the direction of u u = v v Finally calculate v 3 = a 3 < v a 1 3 > v 1 The components of a 3 in the direction of v 1 and v are taken out. That is substract < v a 3 > v So the corresponding orthonormal set is given by, e i = v i v i 5. Consider a basis {e 1, e } = {( 1 ), ( )} to represent a vector v = (3 ) in a two 3 5 dimensional space such that, v = c 1 e 1 + c e Find a x matrix [A], such that v = [A]. u where u = ( c 1 c ) Solution: First let us find u c 1 ( 1 ) + c ( 3 ) = (3 5 ) Joint initiative of IITs and IISc Funded by MHRD Page 7 of 8
c 1 c = 3 c 1 + 3c = 5 } solving for c 1 and c c 1 4c = 6 c 1 + 3c = 5 Thus u = ( 19 7 1 7 ) Using [A]. u = v [A] = ( 1 3 ) 7c = + 1 c = 1 7 ; c 1 = 19 7 Joint initiative of IITs and IISc Funded by MHRD Page 8 of 8